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  "body": "\n## Proof\n\nWe argue by contradiction, through the alternating group.\n\n1. Let $n \\geq 5$. The [[alternating group|0T13HM377F]] $A_n$ is the subgroup of [[even permutations|0SQTMQTSRZ]] of the [[symmetric group|0RJMNEN3BJ]] $S_n$.\n2. We compose permutations from right to left. Then $(1\\,2\\,3) = (1\\,3)(1\\,2)$ and $(3\\,4\\,5) = (3\\,5)(3\\,4)$. So both $3$-cycles are even and lie in $A_n$; here we use $n \\geq 5$.\n3. Let $a = (1\\,2\\,3)$ and $b = (3\\,4\\,5)$. Then $ab$ sends $3$ to $a(4) = 4$, while $ba$ sends $3$ to $b(1) = 1$. So $ab \\neq ba$, and $A_n$ is not abelian.\n4. By the [[simplicity of the alternating group|0ARPRJQWZ2]], $A_n$ is [[simple|0C093D1T4M]].\n5. Suppose that $A_n$ is [[solvable|0P3KZK0FTV]]. Take a [[subnormal series|096HN851YZ]] $\\{e\\} = G_0 \\trianglelefteq \\cdots \\trianglelefteq G_k = A_n$ with abelian factors.\n6. Since $A_n \\neq \\{e\\}$, some $G_j$ equals $A_n$ with $j \\geq 1$. Let $j$ be the least such index. Then $G_{j-1} \\neq A_n$.\n7. The subgroup $G_{j-1}$ is normal in $G_j = A_n$. Since $A_n$ is simple and $G_{j-1} \\neq A_n$, we get $G_{j-1} = \\{e\\}$.\n8. So the factor $G_j/G_{j-1} = A_n/\\{e\\}$ is abelian. The map $x \\mapsto x\\{e\\}$ is an isomorphism from $A_n$ onto $A_n/\\{e\\}$, so $A_n$ is abelian.\n9. This contradicts step 3. So $A_n$ is not solvable.\n10. If $S_n$ were solvable, then its subgroup $A_n$ would be solvable, because [[subgroups of solvable groups are solvable|05Y0F9M0TS]]. This contradicts step 9.\n11. Hence $S_n$ is not solvable.\n\n## Idea\n\nA simple group that is not abelian cannot be solvable: any series must jump from the trivial group straight to the whole group, and that factor is not abelian. For $n \\geq 5$ the group $A_n$ is such a group. A solvable $S_n$ would make its subgroup $A_n$ solvable.\n",
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